English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Beliebt Trigonometrie >

2cos^2(2β)+cos(2β)-1=0,-360<= β<= 360

Lösung

2 cos^{2} (2 β) + cos (2 β) - 1 = 0, - 36 0^{\circ} \leq β \leq 36 0^{\circ}

Lösung

β = 3 0^{\circ}, β = 15 0^{\circ}, β = 21 0^{\circ}, β = 33 0^{\circ}, β = - 15 0^{\circ}, β = - 3 0^{\circ}, β = - 33 0^{\circ}, β = - 21 0^{\circ}, β = 9 0^{\circ}, β = 27 0^{\circ}, β = - 9 0^{\circ}, β = - 27 0^{\circ}

Radianten

β = \frac{π}{6}, β = \frac{5 π}{6}, β = \frac{7 π}{6}, β = \frac{11 π}{6}, β = - \frac{5 π}{6}, β = - \frac{π}{6}, β = - \frac{11 π}{6}, β = - \frac{7 π}{6}, β = \frac{π}{2}, β = \frac{3 π}{2}, β = - \frac{π}{2}, β = - \frac{3 π}{2}

Schritte zur Lösung

2 cos^{2} (2 β) + cos (2 β) - 1 = 0, - 36 0^{\circ} \leq β \leq 36 0^{\circ}

Löse mit Substitution

2 cos^{2} (2 β) + cos (2 β) - 1 = 0

Angenommen:

cos (2 β) = u

2 u^{2} + u - 1 = 0

2 u^{2} + u - 1 = 0 : u = \frac{1}{2}, u = - 1

2 u^{2} + u - 1 = 0

Löse mit der quadratischen Formel

2 u^{2} + u - 1 = 0

Quadratische Formel für Gliechungen:

Für

a = 2, b = 1, c = - 1

u_{1, 2} = \frac{- 1 \pm 1 ^{2} - 4 \cdot 2 ( - 1 )}{2 \cdot 2}

u_{1, 2} = \frac{- 1 \pm 1 ^{2} - 4 \cdot 2 ( - 1 )}{2 \cdot 2}

1^{2} - 4 \cdot 2 (- 1) = 3

1^{2} - 4 \cdot 2 (- 1)

Wende Regel an

1^{a} = 1

1^{2} = 1

= 1 - 4 \cdot 2 (- 1)

Wende Regel an

- (- a) = a

= 1 + 4 \cdot 2 \cdot 1

Multipliziere die Zahlen:

4 \cdot 2 \cdot 1 = 8

= 1 + 8

Addiere die Zahlen:

1 + 8 = 9

= 9

Faktorisiere die Zahl:

9 = 3^{2}

= 3^{2}

Wende Radikal Regel an:

n a^{n} = a

3^{2} = 3

= 3

u_{1, 2} = \frac{- 1 \pm 3}{2 \cdot 2}

Trenne die Lösungen

u_{1} = \frac{- 1 + 3}{2 \cdot 2}, u_{2} = \frac{- 1 - 3}{2 \cdot 2}

u = \frac{- 1 + 3}{2 \cdot 2} : \frac{1}{2}

\frac{- 1 + 3}{2 \cdot 2}

Addiere/Subtrahiere die Zahlen:

- 1 + 3 = 2

= \frac{2}{2 \cdot 2}

Multipliziere die Zahlen:

2 \cdot 2 = 4

= \frac{2}{4}

Streiche die gemeinsamen Faktoren:

2

= \frac{1}{2}

u = \frac{- 1 - 3}{2 \cdot 2} : - 1

\frac{- 1 - 3}{2 \cdot 2}

Subtrahiere die Zahlen:

- 1 - 3 = - 4

= \frac{- 4}{2 \cdot 2}

Multipliziere die Zahlen:

2 \cdot 2 = 4

= \frac{- 4}{4}

Wende Bruchregel an:

\frac{- a}{b} = - \frac{a}{b}

= - \frac{4}{4}

Wende Regel an

\frac{a}{a} = 1

= - 1

Die Lösungen für die quadratische Gleichung sind:

u = \frac{1}{2}, u = - 1

Setze in

u = cos (2 β)

ein

cos (2 β) = \frac{1}{2}, cos (2 β) = - 1

cos (2 β) = \frac{1}{2}, cos (2 β) = - 1

cos (2 β) = \frac{1}{2}, - 36 0^{\circ} \leq β \leq 36 0^{\circ} : β = 3 0^{\circ}, β = 15 0^{\circ}, β = 21 0^{\circ}, β = 33 0^{\circ}, β = - 15 0^{\circ}, β = - 3 0^{\circ}, β = - 33 0^{\circ}, β = - 21 0^{\circ}

cos (2 β) = \frac{1}{2}, - 36 0^{\circ} \leq β \leq 36 0^{\circ}

Allgemeine Lösung für

cos (2 β) = \frac{1}{2}

cos (x)

Periodizitätstabelle mit

36 0^{\circ} n

Zyklus:

x 0 3 0^{\circ} 4 5^{\circ} 6 0^{\circ} 9 0^{\circ} 12 0^{\circ} 13 5^{\circ} 15 0^{\circ} cos (x) 1 \frac{3}{2} \frac{2}{2} \frac{1}{2} 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{2} x 18 0^{\circ} 21 0^{\circ} 22 5^{\circ} 24 0^{\circ} 27 0^{\circ} 30 0^{\circ} 31 5^{\circ} 33 0^{\circ} cos (x) - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2} 0 \frac{1}{2} \frac{2}{2} \frac{3}{2}

2 β = 6 0^{\circ} + 36 0^{\circ} n, 2 β = 30 0^{\circ} + 36 0^{\circ} n

2 β = 6 0^{\circ} + 36 0^{\circ} n, 2 β = 30 0^{\circ} + 36 0^{\circ} n

Löse

2 β = 6 0^{\circ} + 36 0^{\circ} n : β = 3 0^{\circ} + 18 0^{\circ} n

2 β = 6 0^{\circ} + 36 0^{\circ} n

Teile beide Seiten durch

2

2 β = 6 0^{\circ} + 36 0^{\circ} n

Teile beide Seiten durch

2

\frac{2 β}{2} = \frac{6 0 ^{\circ}}{2} + \frac{36 0 ^{\circ} n}{2}

Vereinfache

\frac{2 β}{2} = \frac{6 0 ^{\circ}}{2} + \frac{36 0 ^{\circ} n}{2}

Vereinfache

\frac{2 β}{2} : β

\frac{2 β}{2}

Teile die Zahlen:

\frac{2}{2} = 1

= β

Vereinfache

\frac{6 0 ^{\circ}}{2} + \frac{36 0 ^{\circ} n}{2} : 3 0^{\circ} + 18 0^{\circ} n

\frac{6 0 ^{\circ}}{2} + \frac{36 0 ^{\circ} n}{2}

\frac{6 0 ^{\circ}}{2} = 3 0^{\circ}

\frac{6 0 ^{\circ}}{2}

Wende Bruchregel an:

\frac{\frac{b}{c}}{a} = \frac{b}{c \cdot a}

= \frac{18 0 ^{\circ}}{3 \cdot 2}

Multipliziere die Zahlen:

3 \cdot 2 = 6

= 3 0^{\circ}

\frac{36 0 ^{\circ} n}{2} = 18 0^{\circ} n

\frac{36 0 ^{\circ} n}{2}

Teile die Zahlen:

\frac{2}{2} = 1

= 18 0^{\circ} n

= 3 0^{\circ} + 18 0^{\circ} n

β = 3 0^{\circ} + 18 0^{\circ} n

β = 3 0^{\circ} + 18 0^{\circ} n

β = 3 0^{\circ} + 18 0^{\circ} n

Löse

2 β = 30 0^{\circ} + 36 0^{\circ} n : β = 15 0^{\circ} + 18 0^{\circ} n

2 β = 30 0^{\circ} + 36 0^{\circ} n

Teile beide Seiten durch

2

2 β = 30 0^{\circ} + 36 0^{\circ} n

Teile beide Seiten durch

2

\frac{2 β}{2} = \frac{30 0 ^{\circ}}{2} + \frac{36 0 ^{\circ} n}{2}

Vereinfache

\frac{2 β}{2} = \frac{30 0 ^{\circ}}{2} + \frac{36 0 ^{\circ} n}{2}

Vereinfache

\frac{2 β}{2} : β

\frac{2 β}{2}

Teile die Zahlen:

\frac{2}{2} = 1

= β

Vereinfache

\frac{30 0 ^{\circ}}{2} + \frac{36 0 ^{\circ} n}{2} : 15 0^{\circ} + 18 0^{\circ} n

\frac{30 0 ^{\circ}}{2} + \frac{36 0 ^{\circ} n}{2}

\frac{30 0 ^{\circ}}{2} = 15 0^{\circ}

\frac{30 0 ^{\circ}}{2}

Wende Bruchregel an:

\frac{\frac{b}{c}}{a} = \frac{b}{c \cdot a}

= \frac{90 0 ^{\circ}}{3 \cdot 2}

Multipliziere die Zahlen:

3 \cdot 2 = 6

= 15 0^{\circ}

\frac{36 0 ^{\circ} n}{2} = 18 0^{\circ} n

\frac{36 0 ^{\circ} n}{2}

Teile die Zahlen:

\frac{2}{2} = 1

= 18 0^{\circ} n

= 15 0^{\circ} + 18 0^{\circ} n

β = 15 0^{\circ} + 18 0^{\circ} n

β = 15 0^{\circ} + 18 0^{\circ} n

β = 15 0^{\circ} + 18 0^{\circ} n

β = 3 0^{\circ} + 18 0^{\circ} n, β = 15 0^{\circ} + 18 0^{\circ} n

Lösungen für den Bereich

- 36 0^{\circ} \leq β \leq 36 0^{\circ}

β = 3 0^{\circ}, β = 15 0^{\circ}, β = 21 0^{\circ}, β = 33 0^{\circ}, β = - 15 0^{\circ}, β = - 3 0^{\circ}, β = - 33 0^{\circ}, β = - 21 0^{\circ}

cos (2 β) = - 1, - 36 0^{\circ} \leq β \leq 36 0^{\circ} : β = 9 0^{\circ}, β = 27 0^{\circ}, β = - 9 0^{\circ}, β = - 27 0^{\circ}

cos (2 β) = - 1, - 36 0^{\circ} \leq β \leq 36 0^{\circ}

Allgemeine Lösung für

cos (2 β) = - 1

cos (x)

Periodizitätstabelle mit

36 0^{\circ} n

Zyklus:

x 0 3 0^{\circ} 4 5^{\circ} 6 0^{\circ} 9 0^{\circ} 12 0^{\circ} 13 5^{\circ} 15 0^{\circ} cos (x) 1 \frac{3}{2} \frac{2}{2} \frac{1}{2} 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{2} x 18 0^{\circ} 21 0^{\circ} 22 5^{\circ} 24 0^{\circ} 27 0^{\circ} 30 0^{\circ} 31 5^{\circ} 33 0^{\circ} cos (x) - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2} 0 \frac{1}{2} \frac{2}{2} \frac{3}{2}

2 β = 18 0^{\circ} + 36 0^{\circ} n

2 β = 18 0^{\circ} + 36 0^{\circ} n

Löse

2 β = 18 0^{\circ} + 36 0^{\circ} n : β = 9 0^{\circ} + 18 0^{\circ} n

2 β = 18 0^{\circ} + 36 0^{\circ} n

Teile beide Seiten durch

2

2 β = 18 0^{\circ} + 36 0^{\circ} n

Teile beide Seiten durch

2

\frac{2 β}{2} = 9 0^{\circ} + \frac{36 0 ^{\circ} n}{2}

Vereinfache

β = 9 0^{\circ} + 18 0^{\circ} n

β = 9 0^{\circ} + 18 0^{\circ} n

β = 9 0^{\circ} + 18 0^{\circ} n

Lösungen für den Bereich

- 36 0^{\circ} \leq β \leq 36 0^{\circ}

β = 9 0^{\circ}, β = 27 0^{\circ}, β = - 9 0^{\circ}, β = - 27 0^{\circ}

Kombiniere alle Lösungen

β = 3 0^{\circ}, β = 15 0^{\circ}, β = 21 0^{\circ}, β = 33 0^{\circ}, β = - 15 0^{\circ}, β = - 3 0^{\circ}, β = - 33 0^{\circ}, β = - 21 0^{\circ}, β = 9 0^{\circ}, β = 27 0^{\circ}, β = - 9 0^{\circ}, β = - 27 0^{\circ}

Graph

Beliebte Beispiele

cos(2x)= 3/7

cos (2 x) = \frac{3}{7}

0=1-3cos(θ)

0 = 1 - 3 cos (θ)

(tan((3a)/2))tan(a/2)=3

(tan (\frac{3 a}{2})) tan (\frac{a}{2}) = 3

1=sqrt(3)sin(x)

1 = 3 sin (x)

(sin(115))/(20)=(sin(B))/(15)

\frac{sin ( 11 5 ^{\circ} )}{20} = \frac{sin ( B )}{15}